Question

найти наибольшее и наименьшее значение

Answer 1

$y = 2 {x}^{3} +3 {x}^{2} \\ y' = 2 \times 3 {x}^{3 - 1} + 3 \times 2 {x}^{2 - 1} = 6 {x}^{2} + 6x = \\ = 6( {x}^{2} + x) = 6x(x + 1) \\ x_{1} = 0 \: \: \: \: \: \: \: \: x_{2} = - 1 \\ + + + + [ - 1] - - - - [0] + + + + \\ x_{max} = - 1 \\x _{min} = 0 \\ \\ y( - 1) = 2 \times ( - 1) {}^{3} + 3 \times ( - 1) {}^{2} = 2 \times ( - 1) + 3 \times 1 = \\ = - 2 + 3 = 1 \\ \\ y(0) = 2 \times {0}^{3} + 3 \times {0}^{2} = 0 \\ \\ y(1) = 2 \times {1}^{3} + 3 \times {1}^{2} = 2 + 3 = 5$

Ответ: у max = 5 ; y min = 0